solve by elimination

frctl

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5x - 7y = 2
9x - 3y = 6

isolating in eqn1
9x = 6 + 3y
x = 6/9 + 3/9
x = 2/3 + 1/3y

isolating in eqn2
5x = 2 + 7y
x = 2/5 + 7/5y

If I substitute either into the other eqn it is difficult.
Can someone help me arrive at the solutions?
 

Jomo

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Just solve 2/5 + 7y/5 = 2/3 + 1y/3
I would start by multiply both sides by 15 to get rid of all denominators.
Please continue.
 

MarkFL

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If I were instructed to solve by elimination, I would look for the LCM of the corresponding coefficients of either variable. Suppose we wish to eliminate \(y\)...we see the corresponding coefficients are relatively prime, and so the LCM will be their product, 21. So I would multiply the first equation by 3 and the second equation by -7 so that the coefficients on the \(y\) terms will be of equal magnitude, but opposite in sign:

\(\displaystyle 15x-21y=6\)

\(\displaystyle -63x+21y=-42 \)

Now, we can add the two equations and eliminate \(y\):

\(\displaystyle -48x=-36\implies x=\frac{3}{4}\)

Now we can substitute for \(x\) into either original equation to find \(y\).
 

Otis

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solve by elimination …
x = 2/3 + 1/3y
x = 2/5 + 7/5y
Hi. You have two different expressions for the number x (each of them in terms of y). Those expressions must be equal. Write the equation:

2/3 + 1/3y = 2/5 + 7/5y

Now you have an equation that contains only y. In other words, symbol x has been eliminated. That's why it's called the elimination method.

Using Jomo's suggestion (multiply each side by the LCM of 3 and 5), we get:

10 + 5y = 6 + 21y

Solve that for y. Once you have y, you can substitute it into one of your expressions for x above, like 2/3+(1/3)(y).

What textbook are you using?

🤔
 
Last edited:

frctl

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I am using Linear Algebra Step by Step.

Following MarkFL's instructions:
multiply the first equation by 3 and the second equation by -7
15x - 21y = 6
-63x + 21y = -42
addition of both equations eliminates y
-48x = -36
x = 3/4
substitute x in equation1
Which is the original equation I must substitute x into
I obtain fractions such as
15/4 - 7y = 6

I don't see how you obtained 10 + 5y = 6 + 21y
 

Otis

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I obtain fractions such as
15/4 - 7y = 2
That's because x is a fraction. Now solve the equation for y. If you need to clear the fraction first, multiply each side by 4.

\(\;\)
 

frctl

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multiplying both sides by 4
60/4 - 28y = 24
 

frctl

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15 - 28y = 24
-28y = 9

this fraction does not work
I must have made a mistake
 
Last edited:

Otis

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15 - 28y = 24 …
The left-hand side is correct.. The right-hand side is not correct.

Double-check your earlier arithmetic, when you multiplied by 4 on the right.

\(\;\)
 

frctl

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15 - 28y = 8
-28y = -7
y = 1/4
 

Otis

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Good. That equation is correct.

It seems like it's been a long time since you've had to do arithmetic and basic algebra by hand. Is that right? I'm wondering also why you're trying to teach yourself linear algebra topics. Is it for fun?

:)
 

frctl

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Thank you, I would like to take a course, I have many other questions.
 

Otis

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If taking a linear algebra course is your goal, then my suggestion is to review basic algebra and working with ratios (before you enroll). You'll probably need to do some arithmetic by hand on tests.

x = 3/4

y = ?

\(\;\)
 

Otis

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… I don't see how you obtained 10 + 5y = 6 + 21y …
I missed your edit above. We had set your two initial expressions for x equal to one another.

2/3 + 1/3y = 2/5 + 7/5y

I multiplied each side by 15 (the LCM of the denominators), to clear the fractions. In other words, 15/3 and 15/5 become 5 and 3, respectively.

\(\displaystyle \frac{\cancel{15}^5}{1} \cdot \frac{2}{\cancel{3}_1} \;\; + \;\; \frac{\cancel{15}^5}{1} \cdot \frac{y}{\cancel{3}_1} \;\; = \frac{\cancel{15}^3}{1} \cdot \frac{2}{\cancel{5}_1} \;\; + \;\; \frac{\cancel{15}^3}{1} \cdot \frac{7y}{\cancel{5}_1}\)

\(\displaystyle 5 \cdot 2 \;\; + \;\; 5 \cdot y \;\; = \;\; 3 \cdot 2 \;\; + \;\; 3 \cdot 7y\)

That's 10 + 5y = 6 + 21y

There are a lot of different ways to work with fractions. Did you check out the fractions link I'd posted in another thread?

\(\;\)
 

Jomo

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Good. That equation is correct.

It seems like it's been a long time since you've had to do arithmetic and basic algebra by hand. Is that right? I'm wondering also why you're trying to teach yourself linear algebra topics. Is it for fun?

:)
Isn't Linear Algebra always fun?!
 
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